Hill Stat Reviews Exercises and Solutions

60

Iprurarnlete:r, m,

relative to out than the var(t(m)) = approaches

be used to back of this

variables and if VI and

(B.35)

B.6 EXERCISES 497

B.6 Exercises

B.1 * Let X be a discrete random variable with values x = 0, 1,2 and probabilities P(X = 0) = 0.25, P(X = 1) = 0.50, and P(X = 2) = 0.25, respectively. (a) Find E(X). (b) Find E(X2). (c) Find var(X). (d) Find the expected value and variance of g(X) = 3X + 2.

20

498 REVIEW OF PROBABILITY CONCEPTS

B,2 Let X be a discrete random variable that is the value shown on a single roll of a fair die. (a) Represent the probability density function f(x) in tabular form. (b) What is the probability that X = 4? That X = 4 or X = 5? (c) What is the expected value of X? Explain the meaning of E(X) in this case. (d) Find the expected value of X2. (e) Find the variance of X. (f) Obtain a die. Roll it 20 times and record the values obtained. What is the average

of the first 5 values? The first 1O? What is the average of the 20 rolls?

B,3 Let X be a continuous random variable whose probability density function is

{2x o<x< 1

f(x) = ° othenvtse

(a) Sketch the probability density function f(x). (b) Geometrically calculate the probability that X falls between ° and 1/2. (c) Geometrically calculate the probability that X falls between 1/4 and 3/4.

B.4 Let the binary random variable X have pdff(x) = ]7"{1 - p) I-x for x = 0, 1. (a) Find the mean and variance of X. (b) Let XI,." ,Xn be independent discrete (0,1) random variables each with

probability density function f{x). The random variable B Xl + X2 + ... + Xn has a binomial distribution with parameters nand p. The values of Bare b = 0, ... ,n and represent the number of "successes" (i.e., Xi = 1) in n independent trials of an experiment, each with probability p of success. Calculate the mean and variance of B using the rules of expected values and variances. [Hint: What is the expected value of a sum? The variance?]

(c) Let XI , ... ,Xn be independent discrete (0, 1) random variables with probability density functions f{x). The random variable B = XI + X2 + ... + Xn has a binomial distribution. The random variable Y = B / n is the proportion of successes in n trials of an experiment. Find the mean and variance of Yusing the rules of expected values and variances.

B,5 The joint probability density function of two discrete random variables X and Yis given by the following table:

x 2

4

6

1/8 1/4 1/8

y

3 9

1/24 1/4 1/24

1/12 o

1/12

(a) Find the marginal probability density function of Y. (b) Find the conditional probability density function of Y given that X = 2. (c) Find the covariance of X and Y. (d) Are X and Y independent?

B,6* Consider the population in Table B,5 and the joint distribution of the random variables X and Y in Table B,8. described in Section B.3.3. (a) Find the expected value and variance of the random variable 1'. using the

probability distribution in Table B,6.

B.

B

B;

B

21

Nuri Jazairi

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roll of a fair die.

in this case.

is the average rolls?

InunctIOn is

and 1/2. 4 and 3/4.

probability ... +Xn, has a proportion of

of Yusing

X 2.

the random

Y, using the

B.6 EXERCISES 499

(b) Find the covariance between X and Y. (c) Find the correlation between X and Y.

R7 Let XI,X2, ... ,Xn be independent random variables which all have the same probability distribution, with mean fL and variance (12. Let

_ 1 II

X = - IXi ni=l

(a) Use the properties of expected values to show that E(K) fL. (b) Use the properties of variance to show that var(K) (12/n. How have you used

the assumption of independence?

B.8 Suppose that Y1 , Y2 , Y3 is a sample of observations from a N(fL, (12) population but that Y1, Y2 , and Y3 are not independent. In fact suppose

Let Y = (Yl + Y2 + Y3)/3. (a) Find E(Y) (b) Find var(r)

R9* The length of life (in years) of a personal computer is approximately normally distributed with mean 2.9 years and variance 1.96 years. (a) What fraction of computers will fail in the first year? (b) What fraction of computers will last 4 years or more? (c) What fraction of computers will last at least 2 years? (d) What fraction of computers will last more than 2.5 years but less than 4 years? (e) If the manufacturer adopts a warranty policy in which only 5% of the computers

have to be replaced, what will be the length of the warranty period?

RIO Based on long years of experience, an instructor in the principles of economics has determined that in her class the probability distribution of X, the number of students absent on Mondays, is as follows:

x o 2 3 4 5 6 7

f(x) 0.005 0.025 0.310 0.340 0.220 0.080 0.019 0.001

(a) Sketch a probability function of X. (b) Find the probability that on a given Monday either two or three or four students

will be absent. (c) Find the probability that on a given Monday more than three students are absent. (d) Compute the expected value of the random variable X. Interpret this expected

value. (e) Compute the variance and standard deviation of the random variable X. (f) Compute the expected value and variance of Y 7X + 3.

B.ll * Let X be a continuous random variable with probability density function given by

I(x) -O.5x + 1, 0:::; x :::; 2

(a) Graph the density function I(x). (b) Find the total area beneath I (x) for 0 :::; x :::; 2.

22

500 REVIEW OF PROBABILITY CONCEPTS

(c) Find P(X;;:: 1). (d) Find P(X s: 0.5). (e) Find P(X = 1.5).

B.12 Suppose a certain mutual fund has an annual rate of return that is approximately normally distributed with mean (expected value) 10% and standard deviation 4%. Use Table 1 for parts (a)-(c). (a) Find the probability that your I-year return will be negative. (b) Find the probability that your I-year return will exceed 15%. (c) If the mutual fund managers modify the composition of its portfolio, they can

raise its mean annual return to 12%, but this change will also raise the standard deviation of returns to 5%. Answer parts (a) and (b) in light of these decisions. Would you advise the fund managers to make this portfolio change?

(d) Verify your computations in (a)-(c) using your computer software.

B.13* An investor holds a portfolio consisting of two stocks. She puts 25% of her money in Stock A and 75% into Stock B. Stock A has an expected return of RA = 8% and a standard deviation of (J' A 12%. Stock B has an expected return of RB 15% with a standard deviation of (J'B = 22%. The portfolio return is P 0.25RA + 0.75RB. (a) Compute the expected return on the portfolio. (b) Compute the standard deviation of the returns on the portfolio assuming that the

two stocks' returns are perfectly positively correlated. (c) Compute the standard deviation of the returns on the portfolio assuming that the

two stocks' returns have a correlation of 0.5. (d) Compute the standard deviati on of the returns on the portfolio assuming that the

two stocks' returns are uncorrelated.

Ap

R It

Lee

Base

1.

2.

3.

4

5 (j.

1

Ie

L

1

23

Nuri Jazairi

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Nuri Jazairi

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Nuri Jazairi

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C.10 Exercises

C.I Suppose YI , Y2, ••• , YN is a random sample from a population with mean fJ. and variance 0'2. Rather than using all N observations consider an easy estimator of fJ. that uses only the first two observations

Y* =!1 + Yz 2

(~

(t « «

C.2 S

(~

(t « « (f

C.3* T

'" tt CA S

d II

C.5* J! II

i; C e C

C

C.6 .

C.7

67

thus like

to the

and that

(a) Show that Y* is a linear estimator. (b) Show that Y* is an unbiased estimator. (c) Find the variance of Y*.

C.IO EXERCISES 545

(d) Explain why the sample mean of all N observations is a better estimator than Y*.

C.2 Suppose that Y1 , Yz, Y3 is a random sample from a N(I1-, 0-2) population. To estimate 11-consider the weighted estimator

- 1 1 1 Y = - Yt + - Y2 + - Y3 236

(a) Show that Y is a linear estimator. (b) Show that Y is an unbiased estimator. (c) Find the variance of Y and compare it to the variance of the sample mean Y. (d) Is Y as good an estimator as Y? (e) If 0-2 = 9, calculate the probability that each estimator is within I unit on either

side of 11-.

C.3 * The hourly sales of fried chicken at Louisiana Fried Chicken are normally distributed with mean 2000 pieces and standard deviation 500 pieces. What is the probability that in a 9-hour day more than 20,000 pieces will be sold?

C.4 Starting salaries for Economics majors have a mean of $47,000 and a standard deviation of $8,000. What is the probability that a random sample of 40 Economics majors will have an average salary of more than $50,000?

C.5* A store manager designs a new accounting system that will be cost effective if the mean monthly charge account balance is more than $170. A sample of 400 accounts is randomly selected. The sample mean balance is $178 and the sample standard deviation is $65. Can the manager conclude that the new system will be cost effective? (a) Carry out a hypothesis test to answer this question. Use the (t 0.05 level of

significance. (b) Compute the p-value of the test.

C.6 An econometric professor's rule of thumb is that students should expect to spend 2 hours outside of class on coursework for each hour in class. For a 3 hour per week class, this means that students are expected to do 6 hours of work outside class. The professor randomly selects eight students from a class, and asks how many hours they studied econometrics during the past week. The sample values are 1, 3, 4,4,6,6, 8, 12. (a) Assuming that the population is normally distributed, can the professor conclude

at the 0.05 level of significance that the students are studying on average at least 6 hours per week?

(b) Construct a 90% confidence interval for the popUlation mean number of hours studied per week.

C.7 Modern labor practices attempt to keep labor costs low by hiring and laying off workers to meet demand. Newly hired workers are not as productive as experienced ones. Assume assembly line workers with experience handle 500 pieces per day. A manager concludes it is cost effective to maintain the current practice if new hires, with a week of training, can process at least 450 pieces per day. A random sample of N 50 trainees is observed. Let Y; denote the number of pieces each handles on a

68

546 REVIEW OF STATISTICAL INFERENCE

randomly selected day. The sample mean is y = 460 and the estimated sample standard deviation is (j = 38. (a) Carry out a test of whether or not there is evidence to support the conjecture that

current hiring procedures are effective, at the 5% level of significance. Pay careful attention when formulating the null and alternative hypotheses.

(b) What exactly would a Type I error be in this example? Would it be a costly one to make?

(c) Compute the p-value for this test.

C.8* To evaluate alternative retirement benefit packages for its employees, a large corporation must determine the mean age of its workforce. Assume that the age of its employees is normally distributed. Since the corporation has thousands of workers a sample is to be taken. If the standard deviation of ages is known to be (J' 21 years, how large should the sample be to ensure that a 95% interval estimate of mean age is no more than 4 years wide?

C.9 Consider the discrete random variable Y that takes the values y = 1,2, 3, and 4 with probabilities 0.1, 0.2. 0.3, and 0.4, respectively. (a) Sketch this pdf (b) Find the expected value of Y (c) Find the variance of Y (d) If we take a random sample of size N = 3 from this distribution, what are the

mean and variance of the sample mean, ¥ = (Yl + Y2 + Y3)/3?

C.l

69

419

APPENDIX B

Exercise Solutions

Appendix B, Exercise Solutions, Principles of Econometrics, 3e 420

EXERCISE B.1

(a) ( )2

0( ) 0 0.25 1 0.50 2 0.25 1i i

iE X x P X x

== = = × + × + × =∑

(b) ( ) ( )2

2 2 2 2 2

00 0.25 1 0.50 2 0.25 1.5i i

iE X x P X x

== = = × + × + × =∑

(c) ( ) ( ) ( ) ( ) ( )2 2 2 2 2

0var( ) ( ) 0 1 0.25 1 1 0.50 2 1 0.25 0.5i i

iX x E X P X x

== − = = − × + − × + − × =∑

(d) ( ) (3 2) 3 ( ) 2 5E g X E X E X⎡ ⎤ = + = + =⎣ ⎦ ( ) 2var var(3 2) 3 var( ) 4.5g X X X⎡ ⎤ = + = =⎣ ⎦


EXERCISE B.2

(a) Probabilities of a single roll of a fair die

X ( )f x

1 16

2 16

3 16

4 16

5 16

6 16

(b) ( )4 1 6P X = = ( ) ( ) ( )4 or 5 4 5 1 6 1 6 1 3P X X P X P X= = = = + = = + =

(c) 6

1( ) . ( ) 1 1 6 2 1 6 3 1 6 4 1 6 5 1 6 6 1 6 3.5

xE X x f x

== = × + × + × + × + × + × =∑

( ) 3.5E X = is the average face value, if the dice is rolled a large number of times.

(d) 6

2 2 2 2 2 2 2 2

1

91( ) . ( ) 1 1 6 2 1 6 3 1 6 4 1 6 5 1 6 6 1 66x

E X x f x=

= = × + × + × + × + × + × =∑

(e) The variance is given by

( ) ( )

( ) ( ) ( ) ( )( ) ( )

6 2

1

2 2 2 2

2 2

var( ) ( )

1 3.5 1 6 2 3.5 1 6 3 3.5 1 6 4 3.5 1 6

5 3.5 1 6 6 3.5 1 6

2.917

i ii

X x E X P X x=

= − ⋅ =

= − × + − × + − × + − ×

+ − × + − ×

=

∑

(f) Average values after 5 rolls, 10 rolls and 20 rolls.


EXERCISE B.3

(a) A sketch of the density function

( )2 0 10 otherwise

x xf x

≤ ≤⎧= ⎨⎩

is as follows (b) When 1/ 2, ( ) 1x f x= = . The probability (0 0.5)P X≤ ≤ is the area of a triangle with base = 0.5

and height = 1. Since area of a triangle = 0.5 base×height× , the probability is given by

(0 0.5) 0.5 0.5 1 0.25P X≤ ≤ = × × = (c) When 1/ 4, ( ) 2 / 4x f x = = . When 3/ 4, ( ) 6 / 4x f x= = . The probability (0.25 0.75)P X≤ ≤ is

given by the area of a rectangle with base = 0.5 and height = 0.5 plus the area of a triangle with base = 0.5 and height = (1.5 0.5)− . That is,

(0.25 0.75) (0.5 0.5) (0.5 0.5 1) 0.5P X≤ ≤ = × + × × =

f(x)

x 1

2


EXERCISE B.4

(a) Using the pdf ( ) ( )11 xxf x p p −= − for x = 0,1:

The mean of the binary random variable, X, is

( ) ( ) ( ) ( ) ( )1 110 0 1 1 1x

E X x f x f f p p p−= = × + × = − =∑

The variance of X is

( ) ( )( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( )( )( )( )

2 2 2 1

2 21 0 1 10 1

22

var 1

1 1 1

1 1

1 1

1

xx

x xX E X E X x p f x x p p p

p p p p p p

p p p p

p p p p

p p

−

− −

= − = − = − −

= − − + − −

= − + −

= − + −

= −

∑ ∑

(b) 1 2 1 2( ) ( ) ( ) ( ) ( )n nE B E X X X E X E X E X= + + + = + + +

p p p np= + + + = ( ) ( )1 2var var ... nB X X X= + + +

( ) ( ) ( )1 2= var var var nX X X+ + + , since 1,..., nX X are independent

( ) ( ) ( ) ( )1 1 1 1p p p p p p np p= − + − + + − = −

(c) 1( ) ( )B npE Y E E B pn n n

⎛ ⎞= = = =⎜ ⎟⎝ ⎠

( ) ( ) ( ) ( )2 2

1 11var var varnp p p pBY B

n n n n− −⎛ ⎞= = = =⎜ ⎟

⎝ ⎠


EXERCISE B.5

(a) The marginal probability density function of Y is h(y) where

1 1 1 1(1)8 4 8 2

h = + + =

1 1 1 1(3)24 4 24 3

h = + + =

1 1 1(9) 012 12 6

h = + + =

(b) The conditional probability density function for y given 2X = is ( | 2) (2, ) (2)f y X f y g= = ,

where ( )g x is the marginal probability density function for x.

Now,

1 1 1 1(2)8 24 12 4

g = + + =

and so ( | 2)f y X = is given by

(2,1) 1/8 1(1| 2)(2) 1/ 4 2

ff Xg

= = = =

(2,3) 1/ 24 1(3| 2)(2) 1/ 4 6

ff Xg

= = = =

(2,9) 1/12 1(9| 2)(2) 1/ 4 3

ff Xg

= = = =

(c) ( )cov , [ ( )][ ( )]X Y E X E X Y E Y= − −

( ) 1 1 14 2 42 4 6 4E X = × + × + × =

( ) 1 1 12 3 61 3 9 3E Y = × + × + × =

( ) ( )( ) ( )

( )( ) ( )( ) ( )( ) ( )( )1 1 1 18 24 24 12

cov , 4 3 ,

2 4 1 3 2 4 3 3 6 4 3 3 6 4 9 3

0

x yX Y x y f x y= − −

= − − + − − + + − − + − −

=

∑∑

(d) This is an example where X and Y are not independent, despite the fact that their covariance is

zero. If X and Y are independent, then ( , ) ( ) ( )f x y g x h y= . To show that independence does not hold, consider the example when x = 2 and y = 3,

( ) ( ) ( )1 1 1 12,3 2 324 4 3 12

f g h= ≠ = × =


EXERCISE B.6

(a) ( ) 0 0.6 1 0.4 0.4E Y = × + × =

2 2var( ) (0 0.4) 0.6 (1 0.4) 0.4 0.24Y = − × + − × = (b) ( ) 1 0.1 2 0.2 3 0.3 4 0.4 3E X = × + × + × + × =

2 2 2 2var( ) (1 3) 0.1 (2 3) 0.2 (3 3) 0.3 (4 3) 0.4 1X = − × + − × + − × + − × =

( ) 0 0 0.1 0 2 0.1 1 3 0.1 1 4 0.1 1E XY = × × + × × + + × × + × × =

( ) ( ) ( ) ( ) ( ) ( )cov , [ ][ ] 1 3 0.4 0.2X Y E X E X Y E Y E XY E X E Y= − − = − = − × = −

(c) cov( ) 0.2ρ 0.408var( ) var( ) 1 0.24

XYX Y

−= = = −


EXERCISE B.7

(a) ( ) ( ) ( )1 2 1 21 1... ( ) ( ) ( )n nE X E X X X E X E X E Xn n⎡ ⎤= + + + = + + +⎢ ⎥⎣ ⎦

( )1 nn n

μ= μ + μ + + μ = = μ

(b) ( ) ( )1 21var var nX X X Xn

⎛ ⎞= + + +⎜ ⎟⎝ ⎠

( ) ( ) ( )( )1 22

1 var var var nX X Xn

= + + +

2

22

1 nn n

σ= σ =

Since 1 2, ,..., nX X X are independent random variables, their covariances are zero.


EXERCISE B.8

(a) ( ) ( ) ( )3 3

1 1

1 1 1 33 3 3i i

i iE Y E Y E Y

= =

⎡ ⎤= = = μ = μ⎢ ⎥⎣ ⎦∑ ∑

(b) ( ) ( )3

1 2 31

1 1var var var3 9i

iY Y Y Y Y

=

⎛ ⎞= = + +⎜ ⎟⎝ ⎠∑

( ) ( ) ( )( )

( ) ( )( )

1 2 3 1 2 1 3 2 3

2 2

2 2

2

1 var var + var 2cov( , ) 2cov( , ) 2cov( , )91 3 3 2 0.591 13 3

23

Y Y Y Y Y Y Y Y Y= + + + +

= σ + × σ

= σ + σ

σ=


EXERCISE B.9

(a) ( ) ( ) ( )1 2.91 1.357 1 1.357 0.0871.96

P X P Z P Z P Z−⎛ ⎞< = < = < − = − < =⎜ ⎟

⎝ ⎠

(b) ( ) ( ) ( )4 2.94 0.786 1 0.786 0.2161.4

P X P Z P Z P Z−⎛ ⎞≥ = > = > = − < =⎜ ⎟⎝ ⎠

(c) ( ) ( ) ( )2 2.92 0.643 0.643 0.7401.4

P X P Z P Z P Z−⎛ ⎞≥ = > = > − = < =⎜ ⎟⎝ ⎠

(d) ( ) ( )2.5 2.9 4 2.92.5 4 0.286 0.7861.4 1.4

P X P Z P Z− −⎛ ⎞< < = < < = − < <⎜ ⎟⎝ ⎠

( ) ( )

( ) ( )( )0.786 0.286

0.786 1 0.286

0.396

P Z P Z

P Z P Z

= < − < −

= < − − <

=

(e) We want X0 such that P(X < X0) = 0.05. Now, ( 1.645) 0.95P Z < = and hence

( 1.645) 0.05P Z < − =

Thus, it follows that a suitable X0 is such that

0 2.91.6451.4

X −− =

Solving for X0 yields

( )( )0 2.9 1.645 1.4 0.597X = − = (which is approximately 7 months)


EXERCISE B.10

(a) The probability function of X is shown below.

0.0

0.1

0.2

0.3

0.4

FX

0 1 2 3 4 5 6 7

(b) The probability that, on a given Monday, either 2, or 3, or 4 students will be absent is

4

2( ) (2) (3) (4) 0.310 0.340 0.220 0.87

xf x f f f

== + + = + + =∑

(c) The probability that, on a given Monday, more than 3 students are absent is

7

4( ) (4) (5) (6) (7) 0.220 0.080 0.019 0.001 0.32

xf x f f f f

== + + + = + + + =∑

(d) 7

0( ) . ( ) 0 0.005 1 0.025 2 0.310 3 0.340

xE X x f x

== = × + × + × + ×∑

4 0.220 5 0.080 6 0.019 7 0.001

3.066

+ × + × + × + ×

=

Based on information over many Mondays, the average number of students absent on Mondays is 3.066.

(e) 2 2var( ) ( ) [ ( )]X E X E X= −

72 2 2 2 2 2

02 2 2 2

( ) ( ) 0 0.005 1 0.025 2 0.310 3 0.340

4 0.220 5 0.080 6 0.019 7 0.001 = 10.578x

E X x f x=

= = × + × + × + ×

+ × + × + × + ×

∑

2 2var( ) 10.578 (3.066) 1.17764X = σ = − =

2 1.0852.σ = σ = (f) ( ) (7 3) 7 ( ) 3 7 3.066 3 24.462E Y E X E X= + = + = × + =

2var( ) var(7 3) 7 var( ) 49 1.17764 57.704Y X X= + = = × =


EXERCISE B.11

(a) The probability density function is shown below. (b) Total area of the triangle is half the base multiplied by the height; i.e., the area is 0.5 2 1 1× × = (c) When x = 1, ( ) (1) 0.5.f x f= =

Then, ( 1)P X ≥ is given by the area to the right of 1 which is ( 1 0.5 0.251) 0.5P X × × =≥ = (d) When x = 0.5, ( )0.5f = 0.75.

1( 0.5) 1 ( 0.5) 1 1.5 0.75 0.43752

P X P X≤ = − > = − × × =

(e) For a continuous random variable the probability of observing a single point is zero. Thus, ( 1.5) 0P X = = .

( )f x

1

( )f x

0 1 2x


EXERCISE B.12

(a) 2~ (0.10,0.04 )X N

0 0.1( 0) ( 2.5) 1 0.9938 0.00620.04

P X P Z P Z−⎛ ⎞< = < = < − = − =⎜ ⎟⎝ ⎠

(b) 0.15 0.1( 0.15) ( 1.25) 1 0.8944 0.10560.04

P X P Z P Z−⎛ ⎞> = > = > = − =⎜ ⎟⎝ ⎠

(c) Now 2~ (0.12,0.05 )X N

0 0.12( 0) ( 2.4) 1 0.9918 0.00820.05

P X P Z P Z−⎛ ⎞< = < = < − = − =⎜ ⎟⎝ ⎠

0.15 0.12( 0.15) ( 0.6) 1 0.7257 0.27430.05

P X P Z P Z−⎛ ⎞> = > = > = − =⎜ ⎟⎝ ⎠

The calculations show that the probability of a negative return has increased from 0.62% to

0.82%, while the probability of a return greater than 15% has increased from 10.56% to 27.43%. Whether fund managers should or should not change their portfolios depends on their risk preferences.


EXERCISE B.13

(a) ( ) ( ) ( )( ) 0.25 0.75 0.25 0.75A B A BE P E R R E R E R= + = +

0.25 0.75 13.258 15= × + × = (b) ( ) ( )2var var 0.25 0.75P A BP R R= σ = +

( ) ( ) ( )2 20.25 var 0.75 var 2 0.25 0.75 cov ,A B A BR R R R= + + × × ×

Now,

cov( , )1var( ) var( )

A B

A B

R RR R

ρ = =

Hence,

cov( , ) 12 22 264A B A BR R = σ σ = × =

2 2 2 2var( ) 0.25 12 0.75 22 2 0.25 0.75 264 380.25P = × + × + × × × =

380.25 19.5Pσ = = (c) When

cov( , )0.5var( ) var( )

A B

A B

R RR R

ρ = =

cov( , ) 0.5 0.5 12 22 132A B A BR R = ×σ σ = × × =

2 2 2 2var( ) 0.25 12 0.75 22 2 0.25 0.75 132 330.75P = × + × + × × × =

330.75 18.1865Pσ = = (d) When ρ 0= , cov( , ) 0A BR R = , and the variance and standard deviation of the portfolio are

2 2 2 2var( ) 0.25 12 0.75 22 281.25P = × + × =

281.25 16.7705Pσ = =

433

APPENDIX C

Exercise Solutions

Appendix C, Exercise Solutions, Principles of Econometrics, 3e 434

EXERCISE C.1

(a) A linear estimator is one that can be written in the form i ia Y∑ where ai is a constant. Rearranging Y* yields,

2

1 21 2

1

1 1 1*2 2 2 2 i

i

Y YY Y Y Y=

+= = + =∑

Thus, Y* is a linear estimator where 1 2 for 1,2ia i= = and 0ia = for 3,4, ,i N= … . (b) The expected value of an unbiased estimator is equal to the true population mean.

( ) ( ) ( )1 21 2

1 1 1 1*2 2 2 2 2

Y YE Y E E Y E Y+⎛ ⎞= = + = μ + μ = μ⎜ ⎟⎝ ⎠

(c) The variance of Y* is given by

( ) 1 21 2

1 1var * var var2 2 2

Y YY Y Y+⎛ ⎞ ⎛ ⎞= = +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

( ) ( ) ( )2 2

1 2 1 21 1 1 1var var 2 cov ,2 2 2 2

Y Y Y Y⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2

2 21 14 4 2

σ= σ + σ = , since ( )1 2cov , 0Y Y =

(d) The sample mean is a better estimator because it uses more information. The variance of

the sample mean is 2 Nσ which is smaller than 2 2σ when 2N > , thus making it a better estimator than Y*. In general, increasing sample information reduces sampling variation.


EXERCISE C.2

(a) 3

1 2 31

1 1 1 , where are constants for 1,2 and 3.2 3 6 i i i

iY Y Y Y a Y a i

== + + = =∑

(b) ( ) ( ) ( ) ( )1 2 3 1 2 31 1 1 1 1 1 1 1 12 3 6 2 3 6 2 3 6

E Y E Y Y Y E Y E Y E Y⎛ ⎞= + + = + + = μ + μ + μ = μ⎜ ⎟⎝ ⎠

(c) ( ) 1 2 31 1 1var var2 3 6

Y Y Y Y⎛ ⎞= + +⎜ ⎟⎝ ⎠

( ) ( ) ( ) ( ) ( )1 2 3 1 2 2 31 1 1var var var , since cov , cov , 04 9 36

Y Y Y Y Y Y Y= + + = =

2

2 2 21 1 1 74 9 36 18

σ= σ + σ + σ =

The variance of the sample mean is

( )2 2 26var

3 18Y

Nσ σ σ

= = =

which is smaller than the variance of Y . (d) Since ( ) ( )var varY Y> , Y is not as good an estimator asY .

(e) The probability that the estimatorY is within one unit on either side of μ is:

( ) ( ) ( )

[ ]

1 11 1var var var

1 13 3 3 3

0.577 0.577 0.436

YP Y PY Y Y

P Z

P Z

⎡ ⎤− −μ⎢ ⎥⎡ ⎤μ − ≤ ≤ μ + = ≤ ≤⎣ ⎦ ⎢ ⎥⎣ ⎦

⎡ ⎤= − ≤ ≤⎢ ⎥

⎣ ⎦

= − ≤ ≤ =

The probability that the estimatorY is within one unit on either side of μ is:

( ) ( ) ( )

[ ]

1 11 1var var var

1 17 9 18 7 9 18

0.5345 0.5345 0.407

YP Y PY Y Y

P Z

P Z

⎡ ⎤− −μ⎢ ⎥⎡ ⎤μ − ≤ ≤ μ + = ≤ ≤⎣ ⎦ ⎢ ⎥

⎢ ⎥⎣ ⎦

⎡ ⎤= − ≤ ≤⎢ ⎥

× ×⎢ ⎥⎣ ⎦

= − ≤ ≤ =


EXERCISE C.3

Let X be the random variable denoting the hourly sales of fried chicken which is normally distributed; 2(2000,500 )X N∼ .

The probability that in a 9 hour day, more than 20,000 pieces will be sold is the same as

the probability that average hourly sales of fried chicken is greater than 20,000/9 ≈ 2,222 pieces.

22222222

2222 2000 500 9

666 500

[ 1.332] 0.091

XP X PN N

P Z

P Z

P Z

⎡ ⎤− μ −μ⎡ ⎤> = >⎢ ⎥⎣ ⎦ σ σ⎣ ⎦

⎡ ⎤−= >⎢ ⎥

⎣ ⎦

⎡ ⎤= >⎢ ⎥⎣ ⎦

= > =


EXERCISE C.4

Let the random variable X denote the starting salary for Economics majors. Assume it is normally distributed; 2(47000,8000 )X N∼ .

5000050000

50000 470008000 40

[ 2.37] 1 0.9911 0.0089

XP X PN N

P Z

P Z

⎡ ⎤− μ −μ⎡ ⎤> = >⎢ ⎥⎣ ⎦ σ σ⎣ ⎦

⎡ ⎤−= >⎢ ⎥

⎣ ⎦

= > = − =


EXERCISE C.5

(a) We set up the hypotheses 0 : 170H μ ≤ versus 1 : 170H μ > . The alternative is 1 : 170H μ > because we want to establish whether the mean monthly account balance is more than 170.

The test statistic, given H0 is true, is:

( )399170

ˆXt t

N−

=σ

∼

The rejection region is 1.649t ≥ . The value of the test statistic is

178 170 2.46265 400

t −= =

Since 2.462 1.649t = > , we reject H0 and conclude that the new accounting system is cost

effective. (b) (399) (399)2.462 1 2.462 0.007p P t P t⎡ ⎤ ⎡ ⎤= ≥ = − < =⎣ ⎦ ⎣ ⎦


EXERCISE C.6

(a) To decide whether the students are studying on average at least 6 hours per week, we set up the hypotheses 0 : 6H μ = versus 1 : 6H μ > .

The test statistic, given H0 is true, is

( )76

ˆXt t

N−

=σ

∼

( )

( )

8

1

82 2

1

1 1 1 3 4 4 6 6 8 12 5.58 8

1ˆ var ( ) 11.42867

ii

ii

X x

X x X

=

=

= = + + + + + + + =

σ = = − =

∑

∑

5.5 6 0.59811.4286 8

t −= = −

At the 0.05 level of significance, the rejection region is 1.895t > . Since t = −0.598 < 1.895, we do not reject H0 and therefore cannot conclude that, at the

0.05 level of significance, the students are studying more than 6 hours per week (b) A 95% confidence interval for the population mean number of hours studied per week is:

2ˆ

cX tNσ

± = 11.42865.5 1.8958

± = [ ]3.235, 7.765


EXERCISE C.7

(a) To test whether current hiring procedures are effective, we test the hypothesis that 0 : 450H μ ≤ against 1 : 450.H μ > The manager is interested in workers who can process

at least 450 pieces per day. The test statistic, when H0 is true, is

( )49450

ˆXt t

N−

=σ

∼

The value of the test statistic is

460 450 1.86138 50

t −= =

Using a 5% significance level at 49 degrees of freedom, the rejection region is t > 1.677. Since 1.861 > 1.677, we reject 0H and conclude that the current hiring procedures are

effective. (b) A type I error occurs when we reject the null hypothesis but it is actually true. In this

example, a type I error occurs when we wrongly reject the hypothesis that the hiring procedures are effective. This would be a costly error to make because we would be dismissing a cost effective practice.

(c) p-value = ( )(49)1 ( 1.861)P t− < = (1 − 0.9656) = 0.0344


EXERCISE C.8

The interval estimate of a normally distributed random variable is given by

cY z N± ×σ , where cz is the corresponding critical value at a 95% level of confidence.

The length of the interval is therefore ( )2 cz N× ×σ .

To ensure that the length of the interval is less than 4, derive N as follows:

( )

( )

( )

2

2

2 4

2

4

1.96 21 4

423.525

c

c

c

zN

z N

z N

N

N

σ⎛ ⎞× <⎜ ⎟⎝ ⎠

σ <

σ <

× <

<

A sample size of 424 employees is needed.


EXERCISE C.9

(a) A sketch of the pdf is shown below.

.0

.1

.2

.3

.4

.5

1 2 3 4

pdf

y

(b) ( ) ( )

4

11 0.1 2 0.2 3 0.3 4 0.4 3i i

iE Y y P Y y

== = = × + × + × + × =∑

(c) ( ) ( )( ) ( )4 2

1var i i

iY y E Y P Y y

== − =∑

( ) ( ) ( ) ( )2 2 2 21 3 0.1 2 3 0.2 3 3 0.3 4 3 0.4 1= − × + − × + − × + − × =

(d) ( ) 1 2 3 31 2

3 3 3 3Y Y Y YY YE Y E E E E+ +⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞= = + +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠

( ) ( ) ( )1 2 3

1 1 13 3 31 1 13 3 3 33 3 3

E Y E Y E Y= + +

= × + × + × =

( ) 1 2 3var var3

Y Y YY + +⎛ ⎞= ⎜ ⎟⎝ ⎠

( ) ( ) ( )1 2 3

1 1 1var var var9 9 91 1 1 11 1 19 9 9 3

Y Y Y= + +

= × + × + × =

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